Either of these statements can enable finding the probability of Harold sitting in a given seat, i.e. .

Statement 1 outright provides it, while Statement 2 can lead towards it with the relation 

This is a conditional probability problem, which follows the form:

In this case  is the probability of passing the first test,  is the probability of passing the second test,  is then the probability of passing the second test, having passed the first, and  is the probability of having passed both. The probability or percentage of students that passed the first test is necessary information.

The probability of drawing a red card from a standard deck is .

If a red card is removed from a standard deck of fifty-two cards, and one card is chosen at random, the probability of it being red is . If a black card is removed, the probability is . Therefore, to answer the question, the color of the card must be known; the rank is irrelevant to the question. Statement 1 - but not Statement 2 - provides sufficient information.

The probability of drawing a black card from a standard, unaltered deck is .

If a red card is removed from a standard deck of fifty-two cards and replaced by the joker, and one card is chosen at random, all 26 black cards remain in a deck of 52, so the probability of the chosen card being black is still . If a black card is removed and replaced, there are 25 black cards out of 52 left, so the probability is . Therefore, to answer the question, the color of the card must be known. However, the two statements together, if assumed true, leave two possibilities open - the card could be a diamond (red) or a spade (black). The two statements together provide insufficient information.

Assume both statements to be true. The probability of drawing a black card from a standard, unaltered deck is . We examine two cases.

Case 1: The jack of diamonds from the first deck was replaced by the jack of clubs from the other deck. 

Then there are now 27 black cards in a deck of 52. The probability of drawing a black card at random from this deck has increased from  to .

Case 2: The jack of clubs from the first deck was replaced by the jack of diamonds  from the other deck. 

Then there are now 25 black cards in a deck of 52. The probability of drawing a black card at random from this deck has decreased from  to .

In both scenarios, the conditions of both statements were met; the removed card and its replacement were jacks of different suits. But in one scenario, the probability of drawing a black card increased, and in the other, it decreased. The two statements together provide insufficient information.

The replacement of one card by the joker in a deck of 52 decreases the number of cards of the color of the removed card by 1, leaves as is the number of cards of the other color, and leaves as is the total number of cards. Therefore, the probability that a randomly drawn card will be of the color of the removed card will be reduced. and the probability that it will be of the other color will be the same. 

From Statement 1 alone, since the probability of drawing a red card from the altered deck is the same as that of drawing a red card from an unaltered deck, it follows that the removed card is black. From Statement 2 alone, since the probability of drawing a black card from the altered deck has been reduced, it follows again that the removed card is black. 

Assume Statement 1 alone. There were 26 red cards out of 52 total before the switch, and, since the replacement card was of the same color as the removed card, there were 26 red cards out of 52 after the switch. The probability of drawing a red card stayed the same.

Assume Statement 2 alone. Again, there were 26 red cards out of 52 total before the switch. However, Statement 2 leaves open the possibility of both cards having the same color or different colors, since both black (clubs) and red (hearts and diamonds) cards could have been removed or added. If both cards have the same color, then as in Statement 1, the probability stays the same. But if, for example, the removed card is a club and the added card is a diamond, there are now 27 red cards out of 52, and the probability of drawing a red card has increased to . This makes Statement 2 alone inconclusive.

The probability of drawing an ace (one of thirteen ranks) from an unaltered deck is . From either statement alone, it can be determined that the altered deck has 53 cards, 4 of which are aces; this makes the probability of drawing an ace from this deck . The probability has decreased.

A guitarist grabs a guitar pick out of a dish at random. Find the odds that the pick is green.

I) There are 3 different colors of picks. There are 15 green picks, 45 red picks, and n blue picks

II) There is a total of 143 picks in the dish

Recall that probability can be found by the following:

II) Gives us the total number of outcomes

I) Gives us the desired number of outcomes

So our answer can be found by doing the following:

So there is about a 10.49% chance of getting a green pick.

Don't be distracted by the "n" number of blue picks. We still need II) to find the total number of picks, so both are needed.

The total number of balls in the box will be 

.

Since 

,

it follows that the number of balls is

 .

The number of balls  with a given letter of the alphabet is equal to the number of its position in the alphabet; the probability of a ball with that letter being drawn is that number divided by the total number of balls, 351. Therefore, for this probability to exceed , we must have the relation

.

Therefore, . 

The 11th letter of the alphabet is "K", so in order to answer this question, it suffices to know whether the letter comes after "K" in the alphabet.

The question cannot be answered from either statement alone; both "Barack" and "Obama" include at least one letter that comes after "K" and at least one that does not. However, if both statements are assumed, since both of the letters shared by the words - "A" and "B" - come before "K", the question can be answered in the negative.